# Newbie power supply question

Discussion in 'Electronic Basics' started by JC, May 13, 2004.

1. ### JCGuest

How does a low voltage power supply provide a high amount of current
if its internal resistance is greater then the voltage? (ie. 5V
@>5ohms) I am looking at a circuit that I believe is called a flyback
dc converter, it uses a center tapped secondary for full wave
rectification, and an inductor and capacitor for filtering. For
purposes of my example I assume we can disregard the filter componants
as they are probably in the milli-ohm range, but isn't a typical
schottkey diode around 120 ohms? So if I shorted the output of this 5v
circuit, wouldn't I get 5v/120=.04A. But the spec sheet for the
converter says it can do 30 to 200 amps, how can it do that?

I ask because I am trying to apply what I have been studying by
building some circuits and I thought I would start by building a power
supply. It successfully produces 5 volts, but only a few milli-amps. I
tore apart my project board's 5v 1A wall transformer to see how they
did it. Since they are dealing with 60 cycle AC, they used regular
power diodes, but don't they have an even higher resistance? I know I
can get an amp out of this thing because I tried it.

I have read and re-read everything I have but I am just not getting
it.
Thanks for your help.

2. ### John PopelishGuest

(snip)

At some low current, yes. Diodes are not very well described as a
resistance (ratio of volts per amp) because this ratio changes as the
current changes. A better approximation is that they drop a fairly
fixed voltage over the normal range of current. For Schottkys this is
about a half volt. To get more detailed information, you have to look
at the data sheet for the particular diode. For example, here is the
data sheet for a 1 amp Schottky diode:
http://www.ee.washington.edu/stores/DataSheets/diodes/1n5819.pdf
Note on the Forward Characteristics graph on page 2 that even though
the voltage rises as the current rises, it rises less than in
proportion to the current. It drops about .3 volts at .1 amp, but
only .4 volts at 1 amp and .6 volts at 5 amps.

3. ### Rich GriseGuest

I think a little clarification is in order here - the internal resistance
of a supply is more of a derived value than measured. A power supply
designer
will endeavor to get the lowest internal resistance possible - all it is
is the equivalent of an ordinary resistor in series with a perfect power
supply. Of course, this resistance is probably non-linear.

If your 5V supply _really_ has an internal resistance of 5 ohms, then it's
a crappy supply indeed. Either it can provide 10V at no load and 5V at one
amp, or 5V at no load and 0V at one amp - anything along that load line.

HTH!
Rich

4. ### JCGuest

Thanks for taking time to respond! But I still am not getting it. So
let me describe to you what I have done, maybe you can tell me what I
am not seeing. I looked at several basic schematics on the web and
then wound a primary and a secondary coil. To do a quick test of the
ratios I threw together a bridge rectifier and a filter capacitor on
my proto board. I measured voltage across the outputs and got like
11v. I adjusted down the primary current until I got 5v. I then put a
100 ohm resister across the output and measured the current. I got
something like 7 milli amps (don't have my notes with me)I figured it
out to be as if the 5v was seeing a resistance of over 500 ohms. But
if I measure the output of my proto board's supply in the same way, I
get 5v and 100 ohms gives me .05 amps, just like I would expect. After
reading more to try and figure out what is going on (which I couldn't
find anything) I learned that my diodes may not be fast enough and
that the faster schottkey diodes have a lower forward resistance. So I
got some of those and the current only went up to like 11 milli amps.
That is what led me to take apart the proto board's ac adapter. I
don't understand why when I connect a load to their supply, everything
behaves just like it does in my text books; 5v/100ohms=.05, but when I
do so with mine, it's like I am seeing the resistance of all the
componants within my supply, but I don't see that when I measure the
ac supply in the same way???

5. ### John PopelishGuest

(snip)

Hold it right here. This sentence does not make sense to me. What do
you mean that you actually did when you say that you "adjusted down
the primary current until I got 5v"? Do you have a schematic of the
supply you are having trouble understanding?

6. ### Rich GriseGuest

I got stuck at "then wound a primary and a secondary coil." Do you have
a schematic of what you're trying to assemble? Or even a description?

7. ### JCGuest

Sorry, let me try again at explaining it, that would be faster to type
then an ascii schematic. I am trying to build a power supply a simple
step at a time to get hands on experience. First I made a 555 timer
circuit to drive a primary coil. Then I wound a primary and secondary
coil and put them on a ferrite core. I was shooting for 10v (twice my
5v target output for the supply). To see if I got close (I wound a few
extra turns to be sure to go over rather then under the 10v), I made a
bridge rectifier, and put a capacitor accross the output of the
bridge, and measured the voltage to be 11v. I put a 100 ohm resister
across the output and got I think it was 11 mili-amps. That was not
what I thought I should get. To see what effect it had, I lowered the
current used in the primary so the output was now 5v instead of 11v
and tried the 100 ohm resister again and got 7 milli amps. I am
wondering why I did not get 5v/100 ohms=.05A or 11v/100 ohms=.11A.

Thanks for bearing with me.

8. ### John PopelishGuest

The fog is lifting a bit. I think the problem may be with the drive
circuit. Are you passing DC through the primary or only AC? I really
need to see your circuit configuration. I assume you do not have an
oscilloscope to look at what wave shape is coming out of the
transformer.

The important thing to keep in mind with any inductive component (like
your transformer) is that it must somehow see an average of very
nearly zero volts (except for the resistive drop of the wire) across
each winding over a cycle. The current does not need to average zero
(DC is allowed, under some circumstances) over a cycle, but the
voltage does.